Question

4 Consider the three displacement vectors shown in the figure: Vector A has a magnitude of 8.10 km and a direction that makes an angle θ-290 to the left of the positive y-axis, vector B has a magnitude of 5.50 km and a direction that makes an angle of a -25.0° above the positive x-axis, and vectorC has a magnitude of 4.50 km and a direction that makes an angle β-600° below the negative x-axis. Determine the magnitude of the vector D A + + x 643 km

Answer 1

In component form

A = 8.1 km @ theta = 29

A = -8.1 cos(90 - 29) + 8.1 sin(90 - 29)

A = -3.93 i + 7.08 j

B = 5.5 @ alpha = 25

B = 5.5 cos25 + 5.5 sin25

B = 4.98 i + 2.32 j

C = 4.5 @ beta = 60

C = -4.5 cos60 - 4.5 sin60

C = -2.25 i - 3.89 j

D = A + B + C

D = -3.93 i + 7.08 j + 4.98 i + 2.32 j + -2.25 i - 3.89 j

D = -1.2 i + 5.51 j

D = sqrt (1.2^2 + 5.51^2) = 5.64 km

Hence, D = 5.64 km